## Largest product in a grid (problem 11)

Posted: 12/07/2013 in Project Euler
Tags: , , , , , ,

Problem 11 of PE’s website says:

In the 20×20 grid below, four numbers along a diagonal line have been marked in red.

What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?

I provided a general solution in C. My program allocates dynamically any type of matrix, with any size you want, with any length of adjacent numbers you want the program to analize (this is the only extra data you provide as a parameter of the program from the CLI). It also provides some error management (i.e. if both sizes of the grid are less than the length of the adjacent numbers you want to find the product).

My program works this way:

1. Open your command line interpreter and execute the program providing the length of adjacent numbers you want to evaluate. For example, in this problem it is 4, therefore:

$name_of_program 4 2. Now the program is waiting so that the user provides a matrix to the standard input (the program accepts any size!) and press Enter. In order that the program “catches” that you are done, it is to say, that the EOF of stdin is reached, press CTRL-D (but before of this, don’t forget to press Enter, this step is important). 3. It stores the grid in a new (temporal) file. At this point the width (= the amount of numbers of each line) and the height (= the number of lines of the file) of the grid are calculated and a new grid with that right size is created. 4. All the data stored in the file is saved in the grid. 5. All the required calculations inside the grid are performed. 6. The result is printed out on the standard output and the temporal file created by the program is removed from your computer. Here is the C code: #include <stdio.h> #include <stdlib.h> int **newGrid (unsigned int rows, unsigned int columns); /* It returns a pointer to a 2D (rows x columns) matrix. */ void freeGrid (int **grid, unsigned int columns); /* It frees the memory used for grid. */ void recordStdinToFileOpen (FILE * file); /* Entering user's input is stored in a (temporal) file. */ unsigned int countElementsForGrid (FILE * file, int index_of_array); /* After the grid is saved in a file, this function returns the width and the height of the matrix, depending on whether the value of the "index_of_array" argument is 0 or 1, respectively. */ void recordFileOpenToGrid (FILE * from_file, int **grid, unsigned int width, unsigned int height); /* This functions enters the file we are using to allocate dinamically the data and save it in a grid whose size has been previously calculated. */ unsigned long int getLargestProduct (int **grid, unsigned int width, unsigned int height, unsigned int adjacent); /* This function carries out the largest product of any "adjacent" length of adjacent numbers of the "grid" argument, whose size is "width"x"height", and returns the result. For this problem in particular, adjacent = 4, height = 20, width = 20. */ unsigned long int doTheStuff (FILE * file, unsigned int adjacent); /* It gathers all the functions above and returns the return value of the getLargestProduct() function. */ unsigned long int openFileForGrid (unsigned int adjacent); /* It opens a temporal file, does all the stuff described above, and returns the result of doTheSuff() function. When all the work is done, it removes physically that temporal file of your computer. */ int **newGrid (unsigned int rows, unsigned int columns){ int **grid = NULL; unsigned int j = 0; grid = (int **) malloc (columns * sizeof (int *)); for (j = 0; j < columns; j++){ grid[j] = (int *) malloc (rows * sizeof (int)); } return grid; } void freeGrid (int **grid, unsigned int columns){ int j = 0; for (j = 0; j < columns; j++){ free (grid[j]); } free (grid); } void recordStdinToFileOpen (FILE * file){ int num; fseek (file, 0L, SEEK_SET); while (!feof (stdin)){ fscanf (stdin, "%d", &num); if (!feof (stdin)){ fprintf (file, "%d ", num); if (getchar () == '\n'){ fprintf (file, "\n"); } } } } unsigned int countElementsForGrid (FILE * file, int index_of_array){ unsigned int result = 0; unsigned int i = 0, j = 0; /* During the execution time, press Enter after you paste your grid */ unsigned int measures[2] = {i, j}; int first_line_done = 0; int c; fseek (file, 0L, SEEK_SET); while ((c = fgetc (file)) != EOF){ if ((c == ' ') && first_line_done == 0){ i++; } if (c == '\n'){ j++; first_line_done = 1; } else { continue; } } measures[0] = i; measures[1] = j; result = measures[index_of_array]; return result; } void recordFileOpenToGrid (FILE * from_file, int **grid, unsigned int width, unsigned int height){ int i = 0, j = 0; fseek (from_file, 0L, SEEK_SET); for (j = 0; j < height; j++){ for (i = 0; i < width; i++){ fscanf (from_file, "%d", &grid[j][i]); } } } unsigned long int getLargestProduct (int **grid, unsigned int width, unsigned int height, unsigned int adjacent){ unsigned long int result = 1L; unsigned long int i = 0L, j = 0L, i_temp = i, j_temp = j; unsigned long int max = 1L; for (j = 0L; j < height; j++){ for (i = 0L; i < width; i++){ i_temp = i; j_temp = j; if (width - i_temp >= adjacent){ /* Move left-to-right (horitzontally) */ for (i_temp = i; i_temp < adjacent + i; i_temp++){ result *= (unsigned int) grid[j][i_temp]; } if (result >= max) max = result; result = 1L; } if (height - j >= adjacent){ /* Move up-to-down (vertically) */ for (j_temp = j; j_temp < adjacent + j; j_temp++){ result *= (unsigned int) grid[j_temp][i]; } if (result >= max) max = result; result = 1L; } if ((height - j >= adjacent) && (width - i_temp >= adjacent)){ /* Move up-to-down diagonally (\) */ for (j_temp = j, i_temp = i; (j_temp < adjacent + j) && (i_temp < adjacent + i); j_temp++, i_temp++){ result *= (unsigned int) grid[j_temp][i_temp]; } if (result >= max) max = result; result = 1L; } if ((height - j >= adjacent) && (i_temp >= adjacent - 1)){ /* Move up-to-down diagonally (/) */ for (j_temp = j, i_temp = i; (j_temp < adjacent + j) && (i_temp + 1 > i - adjacent + 1); j_temp++, i_temp--){ result *= (unsigned int) grid [j_temp][i_temp]; } if (result >= max) max = result; result = 1L; } } } result = max; return result; } unsigned long int doTheStuff (FILE * file, unsigned int adjacent){ unsigned long int result = 1; unsigned int width = 0, height = 0; int **grid; recordStdinToFileOpen (file); width = countElementsForGrid (file, 0); height = countElementsForGrid (file, 1); if ((adjacent > width) && (adjacent > height)){ fprintf (stderr, "Error: the parameter must be equal or less than, at least, one side of the grid.\n"); exit (1); } grid = newGrid (width, height); recordFileOpenToGrid (file, grid, width, height); result = getLargestProduct (grid, width, height, adjacent); freeGrid (grid, height); return result; } unsigned long int openFileForGrid (unsigned int adjacent){ unsigned long int escape = 0L; FILE * temp_file; temp_file = fopen ("temp.txt", "w+"); if (temp_file != NULL){ escape = doTheStuff (temp_file, adjacent); } else{ fprintf (stderr, "Error: There were a problem when opening a new file.\n"); exit (1); } fclose (temp_file); system ("rm -r temp.txt"); /* Windows users: replace "rm -r temp.txt" by "DEL temp.txt". */ return escape; } int main (int argc, char *argv[]){ char * endptr; const unsigned int adjacent = (unsigned int) strtol (argv[1], &endptr, 10); unsigned long int result; if (argc < 2 || (int) adjacent <= 0){ fprintf (stderr, "Error: A positive integer as parameter is required.\n"); exit (1); } result = openFileForGrid (adjacent); fprintf (stdout, "\n%lu\n", result); return 0; }  I’m going to show you some examples of its use: $ (gcc -Wall -Wconversion -g prob11.c) && ./a.out 2
1 2
3 4

12
$(gcc -Wall -Wconversion -g prob11.c) && ./a.out 3 1 2 9 1 8 4 9 3 3 7 7 7 648$ (gcc -Wall -Wconversion -g prob11.c) && ./a.out 2
1 6 7 2 3

42


$(gcc -Wall -Wconversion -g prob11.c) && ./a.out 5 3 5 5 3 Error: the parameter must be equal or less than, at least, one side of the grid. $ (gcc -Wall -Wconversion -g prob11.c) && ./a.out -1 Error: A positive integer as parameter is required.

In particular, for this problem:

$(gcc -Wall -Wconversion -g prob11.c) && ./a.out 4 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48 70600674  This is the requested answer. Advertisements ## Summation of primes (problem 10) Posted: 01/07/2013 in Project Euler Tags: , , , Problem 10 of PE’s website says: The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.   Find the sum of all the primes below two million. Note: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. (From Wikipedia) And here is the C code. This time the unique remarkable fact is the fastness of the program: #include <stdio.h> #include <stdlib.h> #include <math.h> int esPrimo (unsigned long long int num); /* Returns 1 if "num" is a prime number. Otherwise it returns 0 */ unsigned long long int sumPrimes (unsigned int below_n); /* Returns the summation of all prime numbers below the number "below_n" */ int esPrimo (unsigned long long int num){ int retorno = 1; unsigned long long int i = 5LL; unsigned long long int r; if (num == 1){ retorno = 0LL; /* 1 is not considered a prime number */ } else if (num < 4) { retorno = 1; } else if (num % 2 == 0) { retorno = 0; } else if (num < 9){ retorno = 1; } else if (num % 3 == 0) { retorno = 0; } else { r = (unsigned long long int) sqrt ((double) num); i = 5LL; while (i <= r){ if (num % i == 0){ retorno = 0; break; } if (num % (i + 2) == 0){ retorno = 0; break; } i = i + 6LL; } } return retorno; } unsigned long long int sumPrimes (unsigned int below_n){ unsigned long long int result = 0; unsigned int i = 0; for (i = 0; i < below_n; i++){ if (esPrimo (i)){ result += i; } } return result; } int main (int argc, char *argv[]){ unsigned int below_number = (unsigned int) atoi (argv[1]); printf ("%llu\n", sumPrimes (below_number)); return 0; }  This time I will execute the program with the Unix time command, which displays the system resource usage of a program during its execution time: $ (gcc -Wall -Wconversion prob10.c) && time ./a.out 2000000 142913828922

 

real 0m1.002s user 0m1.000s sys 0m0.000s

It took only one second to perform the summation of all prime numbers below two million! Take a moment to think about the significance of this fact. What makes this program fast is the efficiency of the algorithm that evaluates whether a number is prime.
Compare it now with the first version of the algorithm I wrote:

int esPrimo (unsigned long long int num){

int retorno = 1;
unsigned long long int i = 2LL;

for (i = 2LL; i <= sqrt ((double) num); i++){
if (num % i == 0){
retorno = 0;
break;
}
}

if (num == 0){
retorno = 0;
}

return retorno;
}


This one is a more straightforward one, because it just implements the primitive definition of prime number. But this is also an inconvenience, because it doesn’t even care about some implications of the definition. For example, the fact that if 2 is not a divisor of, let’s say, $17$, implies that nor 4 neither 6 will be divisors of it, because they can be decomposed by a product where there is at least a $2$. The first algorithm just skips all the composite numbers of the ones evaluated before, and that gives its efficiency.

Let’s measure now the time which this “not optimized” version of the same program spends during its execution time:

(gcc -Wall -Wconversion prob10.c -lm) && time ./a.out 2000000 142913828922

 

real 0m10.278s user 0m10.269s sys 0m0.004s

10x times slower. No more words are needed.

## Special Pythagorean triplet (problem 9)

Posted: 01/07/2013 in Project Euler
Tags: , , ,

Problem 9 of PE’s website says:

A Pythagorean triplet is a set of three natural numbers, $a < b < c$, for which

$a^{2} + b^{2} = c^{2}$

For example, $3^2 + 4^2 = 9 + 16 = 25 = 5^2.$

There exists exactly one Pythagorean triplet for which $a + b + c = 1000.$
Find the product $abc$

First of all, after you submit the correct answer on that website, a new forum is enabled to you so that you can post there the way you reached your solution, and there I found an elegant method that does not require code. I will just quote it:

@theopigott:
"I decided to have a go at this one on paper. I considered Euclid's formula for generating triples (a,b,c): a = m^2 - n^2, b = 2mn, c = m^2 + n^2.

 The aim is that a + b + c = 1000 This implies that 2m^2 + 2mn = 1000, so m^2 + mn = m(m + n) = 500. Now 500 = 2^2 * 3^3, so the possible factorisations (values for m and (m+n)) are: 1 * 500 2 * 250 4 * 125 5 * 100 10 * 50 20 * 25 

However m > n and (m+n) > m, so the only one that would work is m = 20 and m + n = 25, giving n = 5. Substituting back into Euclid's formula gives a = 375, b = 200, c = 425, and the desired product abc = 31875000."

Isn’t that beautiful? 😀 Before of writing my C program to solve the problem, I also tried to do it by hand through Euclid’s formula, but I got stuck at $m(m + n) = 500$. There I had an equation with two variables, so I couldn’t continue (avoiding trial and error). I also tried to solve the system of equations

$a + b + c = 1000 \\ a^{2} + b^{2} = c^{2}$

and performing the change of variables $a=m^2 - n^2$ , $b=2mn$ , $c=m^2+n^2$ (because then I could have a system of two equations with only two variables), but then I got a more complex equation with square roots and everything, so it didn't help to simplify the problem, but right the opposite.

I didn't thought about factorizing 500… shame on me! So instead I wrote the following program that solves the problem by brute force:

#include <stdio.h>
#include <stdlib.h>

double exponenciar (double base, int exponent);
/* Returns the result of (base ^ exponent) */

int findProduct (int sum_numbers);
/* Returns a number a*b*C such that (a, b, c) is a Pythagorean triplet and
a + b + c = sum_number */

double exponenciar (double base, int exponent){

double retorna;

if (exponent == 0){
retorna = 1;
} else {
retorna = base * exponenciar (base, exponent - 1);
}

return retorna;
}

int findProduct (int sum_numbers){

int a = 3, b = 4, c = sum_numbers - a - b;
int square_a = (int) exponenciar (a, 2);
int square_b = (int) exponenciar (b, 2);
int square_c = square_a + square_b;

int result = 0;

for (b = 4; b < sum_numbers; b++){
square_b = (int) exponenciar (b, 2);
for (a = 3; a < sum_numbers; a++){
square_a = (int) exponenciar (a, 2);
square_c = square_a + square_b;
c = 1000 - a - b;

if ((int) exponenciar (a + b + c, 2) == (int) exponenciar (sum_numbers, 2) &&
square_c == (int) exponenciar (c, 2)){
result = a*b*c;
break;
}
}
}

return result;
}

int main (int argc, char *argv[]){

int number = atoi (argv[1]);

printf ("%d\n", findProduct (number));

return 0;
}


Time to compile and to run the program:

## Smallest multiple (problem 5)

Posted: 20/06/2013 in Project Euler
Tags: , , , ,

The problem 5 of PE’s website says:

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

 

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

My first impulse was to solve it by brute force, but that was not really challenging (everyone can do a loop), so I found a more elegant way to find the solution by using some basic number theory facts.

The smallest positive number that is evenly divisible by each number of a given set is just the least common multiple ($lcm$) of all them. You read a bit more here for taking a grasp about what is this.

Let’s be $lcm (a, b)$ and $gcd (a, b)$ the least common multiple and the greatest common divisor (which, as the name says, is the maximum common divisor that a set of numbers have), respectively, of two integers $a$ and $b$. Then the following assertions are true:

a) For any given set of integers $\{n_{1}, ..., n_{m}\}$, the $lcm$ of all them can be carried out by the following recursive formula:

$lcm = (n_{1}, n_{2}, n_3... , n_{m}) = lcm (...lcm(lcm(n_{1}, n_{2}), n_{3}), ... , n_{m})$.

b) $lcm (a, b) = \dfrac{ab}{gcd(a, b)}$

Keeping that in mind, I wrote out three functions for solving the problem:

One function to compute the gcd of a couple of integers and another one for computing the lcm of a couple of integers too. The third one will compute the lcm of an entire set of numbers (expressed in a C array), and I will use the idea expressed in a). Of course, it’s a recursive function whose base case is exactly the $lcm$ of two integers.

#include <stdio.h>
#include <stdlib.h>

long long int gcd (long long int a, long long int b);
/* Carries out the greatest common divisor of "a" and "b" */

long long int lcm (long long int a, long long int b);
/* Carries out the least common multiple of "a" and "b" */

long long int lcmn (long long int nums[], int card);
/* Carries out the least common multiple of the integers of
the "nums[]" array, whose size is "card". */

long long int gcd (long long int a, long long int b){

long long int result;

if (b == 0LL){
result = a;
} else {
result = gcd (b, a % b);
}

return result;
}

long long int lcm (long long int a, long long int b){

long long int leastCommonMultiple = (a*b) / gcd (a, b);

return leastCommonMultiple;
}

long long int lcmn (long long int nums[], int card){

const int args_required_lcm = 2;
long long int current_nums[args_required_lcm];
long long int result;
int iteration = card;

if (iteration == 2){
current_nums[0] = nums[0];
current_nums[1] = nums[1];
} else {
current_nums[1] = nums[iteration - 1];
current_nums[0] = lcmn (nums, iteration - 1);
}

result = lcm (current_nums[0], current_nums[1]);

return result;
}

int main (int argc, char *argv[]){

const int numbers_requested = atoi (argv[1]);
long long int numbers[numbers_requested];
int i = 0;

for (i = 0; i < numbers_requested; i++){
numbers[i] = i + 1LL;
}

printf ("%lld\n", lcmn (numbers, numbers_requested));

return 0;
}


Time to compile and to run the program:

$(gcc -Wall -Wconversion prob5.c) && ./a.out 20 232792560 This is the solution. ## Largest palindrome product (problem 4) Posted: 18/06/2013 in Project Euler Tags: , , , , I’m actually working out the problems proposed in Project Euler’s website and I thought that it could be useful for many people if I post here my solutions for some of them. The problem 4 says: A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 x 99.   Find the largest palindrome made from the product of two 3-digit numbers. I just generalized this problem by writing out a modular program, in C, that finds the largest Palindrome for any number of digits given in the command line. The program may seem larger than necessary, but that’s because I used some general functions I defined a while ago for other purposes, so they work fine with any (valid) input (so at the far end that made me easier to structure the program 🙂 ). #include <stdio.h> #include <stdlib.h> double exponenciar (double base, int exponent); /* Returns the result of (base ^ exponent) */ int calculaXifres (int n); /* Returns the number of digits of the number "n". */ long int aIntegerstoInteger (long int array_de_integers[], int longitud); /* Maps each element of the array of integers "array_de_integers", of "longitud" elements, to each digit of an integer, and returns it back. For instance, aIntegerstoInteger ({2, 1, 5}, 3) will return 215. I will only use this function once, just for making easier the inversion of a number. */ long int invertirNumero (long int num); /* Returns the number "num" inverted */ long int maxValueNdigits (int num_digits); /* Returns the maximum value that a number with "num_digits" digits can reach. For example, maxValueNdigits (4) would return 9999. */ long int largestPalindrome (int num_digits); /* Returns the answer of this problem when num_digits = 3. */ double exponenciar (double base, int exponent){ double retorna; if (exponent == 0){ retorna = 1; } else { retorna = base * exponenciar (base, exponent - 1); } return retorna; } int calculaXifres (int n){ int quocient = n; int comptador = 0; while (quocient > 0){ quocient = quocient / 10; comptador++; } return comptador; } long int aIntegerstoInteger (long int array_de_integers[], int longitud){ int num_elements = longitud; int i = 0; long int integer = 0; for (i = 0; i < num_elements; i++){ integer += array_de_integers[i] * (int) exponenciar (10, num_elements - (i + 1)); } return integer; } long int invertirNumero (long int num){ int longitud = calculaXifres (num); long int num_invertit[longitud]; long int numero_invertit; int i = 0; for (i = 0; i < longitud; i++){ num_invertit[i] = num % 10; num = num / 10; } numero_invertit = aIntegerstoInteger (num_invertit, longitud); return numero_invertit; } long int maxValueNdigits (int num_digits){ int i = 0; long int max_value = 0; for (i = 0; i < num_digits; i++){ max_value = max_value + (int) exponenciar (10, i); } max_value = 9*max_value; return max_value; } long int largestPalindrome (int num_digits){ long int starting = (long int) exponenciar (10, num_digits - 1); long int ending = maxValueNdigits(num_digits); long int i = starting; long int j = i; long int largest_palindrome = 0L; for (i = starting; i <= ending; i++){ for (j = starting; j <= ending; j++){ if (i * j == invertirNumero (i * j) && i * j >= largest_palindrome){ largest_palindrome = i * j; } } } return largest_palindrome; } int main (int argc, char *argv[]){ int num_digits = atoi (argv[1]); printf ("%ld\n", largestPalindrome (num_digits)); return 0; }  Time to compile and to execute the program: $ (gcc -Wall -Wconversion prob4.c) && ./a.out 3 906609

And the three-digit factors are 993 and 913, it is to say, $993 \times 913 = 906609$.